Exercise · ZW-FIRST-ORDER-LINEAR-EQUATION-005 · ★☆☆ Basic
The equation \(y'+2y=4\) has a constant solution. Which one?
Choose one option.
Show answer
- D. \(y=2\)
Explanation
A constant has \(y'=0\), so \(2y=4\) and \(y=2\).
Review the concept: First-order linear equation
A first-order linear equation is a differential equation in which \(y\) and \(y'\) appear to the first power, with a coefficient \(p\) and a free term \(q\) that depend only on \(x\).
\[y'+p(x)\,y=q(x)\]
More practice problems: first-order linear equation
- ZW-FIRST-ORDER-LINEAR-EQUATION-001 Which of these equations is a first-order linear equation?
- ZW-FIRST-ORDER-LINEAR-EQUATION-002 Write y'=3y+x^2 in the form y'+p(x) y=q(x). What is p?
- ZW-FIRST-ORDER-LINEAR-EQUATION-003 Find the general solution of y'+y=0.
- ZW-FIRST-ORDER-LINEAR-EQUATION-004 The equation x y'+2y=x^2, with x>0, is brought to standard form by dividing by x. What is p(x)?
- ZW-FIRST-ORDER-LINEAR-EQUATION-006 Find the general solution of y'+2xy=0.
- ZW-FIRST-ORDER-LINEAR-EQUATION-007 Solve y'+y=0, y(0)=5.
- ZW-FIRST-ORDER-LINEAR-EQUATION-008 Which of these functions is a solution of y'+y=x?
- ZW-FIRST-ORDER-LINEAR-EQUATION-009 The general solution of y'+y=x is y=x-1+Ce^-x. Which is the solution with y(0)=0?
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